1. Field of the Invention
This invention relates generally to apparatus and methods for interfacing time-variant signals, and specifically to the conditioning/simulation of waveforms (such as, for example, blood pressure waveforms) on monitoring equipment.
2. Description of Related Technology
Slowly time-variant waveforms or signals have broad application in many fields, including industry, research, and medicine. Within the medical field, one application of particular interest relates to the monitoring of arterial blood pressure within living subjects (such as human beings and canines). As part of so-called “invasive” arterial blood pressure monitoring, a disposable pressure transducer is commonly utilized to measure the blood pressure of the subject via an invasive arterial line that infuses normal saline at a “KVO” rate into an appropriate blood vessel. KVO (or “keep vein open”) refers to the minimum flow rate required to keep an IV needle from clotting off in the vein. The disposable pressure transducer (DPT) is a passive resistive bridge implemented typically from silicon strain beam technology. The DPT electrically interfaces to a patient monitor available from any number of different vendors such as Hewlett Packard/Agilent/Philips, General Electric/Marquette, Datex-Ohmeda, Datascope/Fakuda-Denshi, Welch Allyn/Protocol, Space Labs, Criticare, Critikon, and Valley Medical, as well as others. The patient monitor supplies the excitation signal to energize the bridge circuit of the DPT, and also provides the signal conditioning of the output derived from the DPT in order to display the subject's blood pressure waveform on the display device of the patient monitor.
The various monitors described above utilize varying methods to energize the bridge and recover the output signal. In one simple scheme (shown as FIG. 1), the bridge 101 of the DPT 100 is driven by a constant direct current (DC) voltage source 102. The value of the DC voltage produced by the source 102 can be any value between 1V and 10V, but is typically set at 5V. This voltage value minimizes the self-heating effect that occurs when resistors R1 104, R2 106, R3 108, R4 110, R5 112, and R6 114 dissipate power. This power dissipation changes the temperature of the bridge sensor resistors R3 108, R4 110, R5 112, and R6 114, and causes their electrical resistance values to change. This change in resistance causes an error in the output of the bridge 101 which is not desired. Usually the manufacturer of the DPT will add additional resistors (not shown) to compensate for such temperature effects, and to calibrate the output of the DPT to a particular value (such as 5 uV/V/mmHg). None-the-less, minimizing the self-heating of the components limits the magnitude of the error, since such electrical compensation is never perfect.
As shown in FIG. 1, the drive signal to the DPT 100 is applied across the +E 120 and −E 122 terminals. Resistors R1 104 and R2 106 limit bridge current, and provide temperature compensation for span. When pressure is applied to the transducer, the geometry of the silicon strain beam is arranged so that resistors R3 108 and R6 114 decrease in value, and the values of resistors R4 110 and R5 112 increase in value by the same amount. This change is typically on the order of 1% of the resistor's resistance value, and is commonly referred to as “Delta R” or ΔR. In a typical configuration, resistors R3 108, R4 110, R5 112, and R6 114 all have the same nominal value referred to as Rb. Resistors R1 and R2 are usually chosen to have equal resistance so that the nominal voltage at terminals +S 130 and −S 132 will be equal to one-half of the drive voltage Ed. For these conditions, the output impedance of the bridge 101 will be Rb, and the input impedance will be given by Eqn. 1:Zin=R1+R2+Rb  (Eqn. 1)Typical values for Zin and Zout (Rb) are 3 K ohms and 300 ohms, respectively. Given these definitions, it can be readily demonstrated that the bridge output voltage across terminals +S 130 and −S 132 can be described by Eqn. 2 as follows:
                    Es        =                ⁢                                            +              S                        -                          (                              -                S                            )                                =                      Ed            ×                          dR              Zin                                                          (                  Eqn          .                                          ⁢          2                )            Derivation of Eqn. 2 is conducted by analyzing an equivalent balanced bridge configuration 200 (FIG. 2) based on the following six assumptions: (i) Resistor R1 202=R2 202; (ii) Resistors R3 204, R4 206, R5 208, and R6 210 are all equal in resistance when the bridge 200 is in balance, and are equal to Rb; (iii) when the bridge unbalances, resistors R3, R4, R5, and R6 all change by an equal amount (dR) as shown in FIG. 3; (iv) the load across terminals +S 230 and −S 232 is an infinite differential impedance; (v) there are no loads between +S and +E or −E; and (vi) there are no loads between −S and +E or −E.
When the bridge in FIG. 2 is in balance, the resistance between terminal 1 240 and terminal 2 242 is given by Rb (i.e., 2×Rb placed in parallel with 2×Rb yields an effective impedance of Rb). In the unbalanced configuration of FIG. 3, the resistance between terminal 1 340 and terminal 2 342 is also given by Rb, since the quantity (R3 +R4+dR−dR)=2Rb, which, when placed in parallel with (R5+R6+dR−dR) 2Rb, yields an effective impedance of Rb. Similarly, the bridge output impedance Zout=Rb. 
The unbalanced circuit of FIG. 3 can be simplified to an equivalent circuit 400 shown as FIG. 4 herein. The input impedance seen by the excitation source Ed is therefore given by Eqn. 3:Zin=R1+R2+Rb  (Eqn. 3)The relationship between Eb (i.e., the equivalent voltage across the bridge) and the excitation voltage Ed is given by Eqn. 4:
                    Eb        =                  Ed          ×                      [                          Rb                                                R                  ⁢                                                                          ⁢                  1                                +                                  R                  ⁢                                                                          ⁢                  2                                +                Rb                                      ]                                              (                  Eqn          .                                          ⁢          4                )            Note, however, that (R1+R2+Rb)=Zin. Hence, Eb can be represented as shown in Eqn. 5:
                    Eb        =                  Ed          ×                      [                          Rb              Zin                        ]                                              (                  Eqn          .                                          ⁢          5                )            The bridge may be analyzed alone, since the voltage Eb across the bridge is constant (and hence Rb is constant). This un-balanced circuit equivalent 500 is shown in FIG. 5 herein. The voltage at node +S 502 is given by Eqn. 6:
                                                                        E                ⁡                                  (                                      +                    S                                    )                                            =                            ⁢                              Eb                ×                                  [                                                            Rb                      +                                              d                        ⁢                                                                                                  ⁢                        R                                                                                    Rb                      +                                              d                        ⁢                                                                                                  ⁢                        R                                            +                      Rb                      -                                              d                        ⁢                                                                                                  ⁢                        R                                                                              ]                                                                                                        =                            ⁢                                                E                  ⁡                                      (                                          +                      S                                        )                                                  =                                  Eb                  ×                                      [                                                                  Rb                        +                                                  d                          ⁢                                                                                                          ⁢                          R                                                                                            2                        ×                        Rb                                                              ]                                                                                                                          =                            ⁢                                                E                  ⁡                                      (                                          +                      s                                        )                                                  =                                ⁢                                  Eb                  ×                                                            [                                                                        1                          2                                                +                                                                              d                            ⁢                                                                                                                  ⁢                            R                                                                                2                            ×                            Rb                                                                                              ]                                        .                                                                                                          (                  Eqn          .                                          ⁢          6                )            A similar expression can be developed for the voltage at node −S 504 (Eqn. 7):
                              E          ⁡                      (                          -              S                        )                          =                  Eb          ×                      [                                          1                2                            -                                                d                  ⁢                                                                          ⁢                  R                                                  2                  ×                  Rb                                                      ]                                              (                  Eqn          .                                          ⁢          7                )            The differential output E(s) is simply E(+s)−E(−s), as follows:
                              E          ⁡                      (            S            )                          =                ⁢                              Eb            ×                          {                                                [                                                            1                      2                                        +                                                                  d                        ⁢                                                                                                  ⁢                        R                                                                    2                        ×                        Rb                                                                              ]                                -                                  [                                                            1                      2                                        -                                                                  d                        ⁢                                                                                                  ⁢                        R                                                                    2                        ×                        Rb                                                                              ]                                            }                                =                                    E              ⁡                              (                S                )                                      =                          Eb              ×                              [                                  dR                  Rb                                ]                                                                        (                  Eqn          .                                          ⁢          8                )            Finally, substituting the result of Eqn. 5 into Eqn. 8 yields Eqn. 2 above:
      E    ⁡          (      s      )        =      Ed    ×          [                        d          ⁢                                          ⁢          R                Zin            ]      
In order for all DPT's to function similarly, they are typically calibrated during manufacture to a standard sensitivity of 5 uV/V/mmHg. This means that for an applied pressure of 100 mmHg, and a drive voltage (Ed)) of 5V, the output Es will be 2.5 mV. This requires that the resistance difference (dR) be 1.5 ohms, according to Eqn. 2 above. For a Zin of 3 k ohms, the full-scale pressure specification for such DPTs is 300 mmHg, which would require a resistance difference (dR) value of 4.5 ohms, and yield an output voltage of 7.5 mV.
It can be shown that the differential resistance dR is a function of pressure:dR=Ks×Zin×P  (Eqn. 9)where:
Ks=scaling factor of 5 uV/V/mmHg, and
P=sensed pressure.
Substituting this result into Eqn. 2 results in the transfer function for the bridge, Eqn. 10:Es=Ed×Ks×P  (Eqn. 10)Note that the output voltage Es is a function of both pressure (P), the input variable, and the drive voltage (Ed) provided by the monitor.
For the 5 VDC drive condition, the signal processing of the output is typically quite limited in scope; e.g., amplifying the bridge output with an instrumentation amplifier, and filtering the output with a 2-pole low pass filter whose cutoff frequency is above any frequency components of interest in the blood pressure signal. A typical valve for such filter cutoff frequency is 45 Hz.
When a user wishes to supply the monitor with and display a waveform other than that derived from the aforementioned DPT (such as that from a digital non-invasive blood pressure monitor such as that manufactured by the Assignee hereof), the DPT must be disconnected from the monitor, and the new signal source electrically substituted. For the case of a fixed DC drive voltage of 5V, a circuit may be readily fashioned to interface the new (e.g. digital) signal source to the patient monitor, such as that shown in FIG. 6.
As illustrated in FIG. 6, resistor R1 602 of the interface circuit 600 allows the patient monitor detect the 3 Kohm impedance value (Zin) it normally sees when using the DPT. Additionally, resistors R2 606 and R3 608 set the differential output impedance (Zout) to 300 ohms. The output of node S+610 is biased to +2.5 V by reference amplifier U1 616, and amplifiers U2 618 and U3 620 (and their associated components) set the −S output according to Eqn. 11:−S=2.506−0.0025×Ein  (Eqn. 11)The differential output between +S and −S is given by Eqn. 12:Es=−0.006+0.0025×Ein  (Eqn. 12)The fixed offset of −6 mV (for the exemplary circuit of FIG. 6) can be “zeroed out” or cancelled by most patient monitors, leaving the output voltage Es a function of the input voltage, and scaled such that a 1V input equals 100 mmHg. An alternative to nulling out the −6 mV with the monitor is to add a zero adjustment to amplifier U2 618.
While the circuit 600 of FIG. 6 (or any other similar circuit) works generally for any monitor that has a fixed +5V DC constant voltage excitation, it has significant shortcomings when one attempts to drive the many different types of patient monitors presently available. Many such patient monitors do not use constant +5V excitation, but rather use bipolar sine wave drive, or even pulse drive as a carrier between 2 KHz and 5 KHz, which is modulated by the pressure signal. These monitors use an AC drive to reduce bridge offset effects, and cancel noise. Furthermore, they require a synchronous demodulator as part of their signal conditioning circuitry, in order to recover the blood pressure modulation signal.
Therefore, in order to drive these monitors with a non-DPT device such as the aforementioned digital NIBPM, a circuit is needed that mimics the electrical profile and operation of a passive transducer bridge. The transfer function of such circuit must be effectively identical to that of the passive bridge, and the input and output impedances must match those of the passive bridge as well. The sensitivity factor of 5 uV/V/mmHg previously described (or any corresponding value for the selected monitor) must also be maintained. The circuit must function with any type of excitation source including constant voltage drive of either polarity, constant current drive of either direction, and any AC voltage or current drive of any waveshape, duty cycle, and DC offset with a frequency of 1 KHz to 10 KHz.
Based on the foregoing, what is needed is an improved apparatus and method for interfacing sensing devices producing time variant waveforms (such as for example the systolic, diastolic, and/or average blood pressure waveforms of a living subject) with monitoring devices. Such apparatus and method would ideally (i) be readily adapted to a variety of different configurations of monitoring or display devices, (ii) have a wide dynamic range; (iii) be able to operate on binary digital input (versus requiring conversion to analog first); (iv) maintain the desired sensitivity factor; (v) function with any type of excitation source including constant voltage drive of either polarity; (vi) function with a constant current drive of either direction, and (vii) function with any AC voltage or current drive of any waveshape, duty cycle, and DC offset. Furthermore, such circuit would optimally be electrically isolable from the monitor, be stable and have minimal error or drift through its simulation range, and include provision for the detecting when the monitoring device was electrically connected thereto.